Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001 CHAPTER 2 DISTRIBUTED-ELEMENT CIRCUIT ANALYSIS TECHNIQUES 2.1 TRANSMISSION LINES A transmission line can be viewed as a multiple combination of small circuit segments shown below in Figure 2.1. The series inductance is due to magnetic field effects and the capacitance is due to electric field coupling between the lines. The losses in the transmission media are depicted by the series and the shunt resistors. These resistors represent the finite conductivity of the conductors and the dielectric insulator between the conductors, respectively. The constants R, G, L, and C are defined as per unit length circuit parameters and the resulting resulting circuit is referred to as a distributed model of a transmission line. The length of the transmission line segment is ∆x . Applying Kirchoff's law to the series and shunt elements respectively, one gets V x,t ( ) L∆ x C∆ x R∆ x G ∆ x I x,t ( ) V x ( ) ∆ x,t I x ( ) ∆ x,t x x ∆ x ∆ x x Figure 2.1 Distributed circuit model for a transmission line.
Transcript
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
CHAPTER
2
DISTRIBUTED-ELEMENT CIRCUIT ANALYSISTECHNIQUES
2.1 TRANSMISSION LINES
A transmission line can be viewed as a multiple combination of small circuit segments shownbelow in Figure 2.1. The series inductance is due to magnetic field effects and the capacitance is due toelectric field coupling between the lines. The losses in the transmission media are depicted by the seriesand the shunt resistors. These resistors represent the finite conductivity of the conductors and the dielectricinsulator between the conductors, respectively. The constants R, G, L, and C are defined as per unit lengthcircuit parameters and the resulting resulting circuit is referred to as a distributed model of a transmissionline. The length of the transmission line segment is ∆x .
Applying Kirchoff's law to the series and shunt elements respectively, one gets
V x,t ( )
L ∆x
C ∆x
R ∆x
G ∆x
I x,t ( )
V + x ( ) ∆x,t
I + x ( ) ∆x,t
x + x ∆ x
∆x
x
Figure 2.1 Distributed circuit model for a transmission line.
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( ) ( ) ( ) ( )t
txIxLtxxIRtxxVtxV
∂∂ ,
,,, ∆+∆=∆+− (2.1)
( ) ( ) ( ) ( )t
txxVxCtxxxVGtxxItxI
∂∂ ,
,,,∆+∆+∆+∆=∆+− (2.2)
Dividing both sides of the above equations by ∆x and taking the limit of both equations as ∆x → 0, results in
( ) ( ) ( )t
txILtxRI
xtxV
∂∂
∂∂ ,
,, +=− (2.3)
( ) ( ) ( )t
txVCtxGV
xtxI
∂∂
∂∂ ,
,, +=− (2.4)
Sinusoidal steady state solutions of the voltage and current can be found by assuming the solutions to be( ) ( ) tjexVtxV ω=, and ( ) ( ) tjexItxI ω=, , i.e., the voltage and current can be described as a phasor which is
a complex vector rotating as a function of time. The amplitude and phase of the phasor is a function of x,the position on the transmission line. Substitution of these for the voltage and current in (2.3) and (2.4)yields
( ) ( )xIZxdxVd −= (2.5)
( ) ( )xVYxdxId −= (2.6)
whereLjRZ ω+= and CjGY ω+=
are known as the distributed impedance and addmittance, respectively. Since the time dependence isremoved from the differential equations one is dealing with only the voltage and current phasors. Furtherdifferentiation of (2.5) and (2.6) resulted in two second order linear differential equations.
( ) ( )xVYZxd
xVd =2
2
(2.7)
( ) ( )xIYZxd
xId =2
2(2.8)
Working with (2.7), the solution must bear the form of e x±γ , where γ, known as the propagation constant is
( )( ) βαωωγ jCjGLjRYZ +=++== (2.9)
The parameter α is known as the attenuation constant and β is called the wave number. The general solutionof the voltage phasor is then
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
where the + and - superscripts are chosen to indicate the propagation direction of the voltage wave. Thecurrent phasor can be derived from (2.10) and (2.5), i.e.
( ) ( )xxxx eVeVLjR
eIeIxI γγγγω
γ −−+−−+ −+
=−= (2.11)
The the forward propagating voltage is related to the forward propagating current characteristic impedancewhich is
CjGLjRLjR
Z c ωω
γω
++=+= (2.12)
If the transmission line is lossless then R=G=0 and in that case the characteristic impedance becomes
CLZ =0 (2.13)
and the propagation constant is given by ( )( ) LCjCjLjYZ ωωωγ +=== 0 implying that
α=0 and LCωβ = (2.14)
The results can be understood by considering a sinusoidal signal propagating in the positive xdirection is given by
( )xtA βω −′cos
Note that for a fixed time that the wave repeats for different x positions along the line separated byintergal multiples of 2π/β. This repeating spacial distance is called the wave length and usually designatedas λ. Therefore, λ=2π/β or the wave number is given by β = 2π/λ. The wave number "β" can be thought ofas a spacial angular frequency analogous to ""ω which is a temporial angular frequency. If time is allowedto advance and then a position for which xt βω − is a constant is called a point of constant phase. Thevelocity of a point of constant phase is called the phase velocity of the wave and equals ω/β = fλ.
The cosine trigometric function can be represented as the real part of a complex number given by
( ) xtjeA βω −′Re
At a fixed x position the complex number, ( )xtjeA βω −′ , is a rotating vector in the complex plane isreferred to a phasor. It is convenient to think of the wave as complex and to omit the "Re" operator. Inanalyzing a circuit if the actual real values are required one only needs to take the real part of the phasor.For the geometry shown in Figure 2.2 the voltage and current for a lossless transmission line are shownbelow where )(A xtje βω −′ represents a forward traveling voltage phasor propagating in the positive x-
direction, while )(B xtje βω +′ represents a reverse traveling phasor, i.e. propagating in the negative x-direction.
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Since the time dependence for a phasor is given by tje ω which is multiplied by a complex numberindependent of time it is convenient to omit it and include it only if the explicit time behavior is required.Therefore a forward and reverse propagating phasor is represented, respectively, by xje β−′A and xje βB′ .
If the explicit time dependence is required one multiplies the phasor by tje ω and then take the real part ofthe resulting function. Using this convention the voltage pahsor on a line is given by
xjxj eex ββ BA)V( ′+′= −
which is consistent with the results of (2.12) with γ = 0+jβ. Since x = L - d, where "d" is the lengthmeasured from the right side of the circuit and "L" is the length of the transmission line.
)()( BA)V( dLjdLj eex −−− ′+′= ββ
djLjdjLj eeeed ββββ −− ′+′= BA)V(
If LjeA β−′= A and LjeB βB′= thendjdj eed ββ −+= BA)V( (2.15)
The current is given by the two phasors
0ZBA
)I(djdj ee
dββ −−= (2.16)
where the propagation constant is
λπωβ 2== LC (2.17)
A forward propagating voltage and current phasor are related by the characteristic impedance, Zo.A reverse propagating voltage and current phasor are related by the negative of the characteristicimpedance, -Zo. The plus sign is required for the forward wave and the negative sign for the reverse waveso that the power associated with the propagating wave has the correct sign. When viewed at an arbitarypoint on the line the power associated with a forward wave would be positive indicating that power appearsto be disipated by the right side of the circuit. On the other hand the power associated with a reverse wave
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
would be negative indicating that power appears to be produced by the right side of the circuit, i.e., the rightside of the circuit appears to be a source for the reverse wave.
2.2 IDEAL TRANSMISSION LINE CIRCUITS
Distributed-element circuits are those where the physical dimensions of one or more of thecomponents affect the circuit performance. In many important circuits the distributed components can berepresented in terms of transmission line models. Therefore, this section will consider the analysis ofcircuits which contain transmission lines as components. Initially lossless transmission lines will beconsidered, since they represent a simpler starting point and the techniques extend naturally to lossy lines.Such lines referred to as ideal transmission lines do not represent a serious restriction since microwave andRF circuits normally use low loss materials and circuit effects are rarely dominated by transmission linelosses.
The analysis of transmission line circuits is a natural extension of lumped-element circuit analysis.Kirchoff's Current Law continues to be true at any node and Kirchoff's Voltage Law must hold around anyloop. Voltage and current relations for lumped element resistances (Ohm's Law), inductive reactances, andcapacitive reactances are augmented with those for transmission lines. Transmission lines are thereforetreated like any other circuit element. Circuits are solved by using the current and voltage relationships tocreate a system of equations which can then be solved for branch currents and node voltages. Transmissionline nodes exist at the input and the output and they may be combined with other elements in series and inparallel. Lumped-element (LE) and distributed-element (DE) circuits are shown below in Figure 2.1 and2.2.
(Z )1
(Z )2
(Z )3
Figure 2.2 An example of a lumped element circuit in which thephysical size and spacing of components are notfactors in determining performance
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Figure 2.3 A distributed element circuit in which transmission lines as well aslumped elements are used as components
A simple distributed element circuit consisting of a source, a transmission line, and a load isillustrated in Figure 2.4. The load and source are assumed to be lumped-element components. It is desiredto find the voltages and currents everywhere in the circuit. At each of the terminals of the transmission lineKirchoff's Laws together with Ohm's law can be applied to obtain Equations (2.18a) and (2.18b).
At the load (d = 0) the reflection coefficient ΓL is the ratio of these two waves, i.e. B/AL =Γ orB A L= Γ and the transmission line voltage and current can be expressed as shown in Equations (2.19a) and(2.19b). Since the source is on the left then we can view the forward propagating voltage wave as a
)A()V( Ldjdj eed ββ −Γ+= (2.19a)
I( ) A( ) ZL 0d e ej d j d= − −β βΓ (2.19b)
stimulus which interacts with the load on the right to produce a reflected wave. At a point "d" along the linea generalized reflection coefficient can be defined as the ratio of reflected voltage, dje β−ΓLA to incident
voltage, djAe β . This reflection coefficient designated Γ can therefore be expressed as Equation (2.20a),and the total transmission line voltage and current is given by Equations (2.20b), and (2.20c).
( ) djLed β2−Γ=Γ=Γ (2.20a)
x d
L
(Z )0
I( )d
V( )d
Z S
VS ZL
θ = βL
Z -L
Z +L Γ =L
Z0
Z0
Figure 2.4 A source, a transmission line, and a load
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
( )Γ+= 1)( djAedV β (2.20b)
( ) 01)( ZAedI dj Γ−= β (2.20c)
At the end of the line "d=0" the voltage is given by ( )Γ+= 1)0( AV
and the current is given by( ) 01)0( ZAI Γ−=
,
which results in a ratio shown in Equation (2.21). The generalized and load reflectioncoefficient have a value that results in the proper voltage current ratio at the load. Equivalently, the loadreflects just the right amount of voltage so that the ratio satisfies the boundary condition (i.e., Kirchoff's &Ohm's Laws) at the output node of the line. The actual magnitude of the voltage and current depends on theconstant "A" which will now be shown to depend upon the input node boundary condition (again Kirchoff'sLaws).
( )( ) L
Lo ZZ
IV =
Γ−Γ+=
11
)0()0( L (2.21a)
0
0
ZZZZ
L
LL +
−=Γ (2.21b)
Specifically at the input of the line )(AL)V(d Lθθ jj ee −Γ+==
and 0L Z)(AL)I(d θθ jj ee −Γ−== .Substitution of these equations into the input Kirchoff's Law condition yields Equation (2.22a). Factoringout the ( )os ZZ + term results in Equation (2.22b) where the term Γs is given by Equation (2.22c)
)(A)(ZA
ZV LL0
ssθθθθ jjjj eeee −− Γ++Γ−= (2.22a)
( )θθ 2
0
0s 1AV j
sLjs ee
ZZZ −ΓΓ−
+= (2.22b)
os
oss ZZ
ZZ+−=Γ (2.22c)
sΓ formally appears as a source reflection coefficient, i.e., the reflection coefficient that would result froma load sZ if an incident wave impinged upon it. It will shortly be shown that such an interpretation isphysically correct. The constant "A" can now be determined as shown below and equals and the voltageand current at an arbitrary point "d" on the transmission line is given by
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While the voltage magnitude is proportional to Vs and depends upon Zs in a complicated way the physical
interpretation of the above results can be seen by using the relationship Λ++++=− 321)1(1 rrrr with
θ2jsL er −ΓΓ= . The voltage expression then becomes
( ) )(1)( 4222
0
0 djL
djjsL
jsL
js
seeeeeV
ZZZ
dV ββθθθ −−−− Γ+×+ΓΓ+ΓΓ+
+= Λ
and
( )( )( )
( )
( )( )( )( )↑↑↑
+−+−+−
−−−
↓↓↓
′′′
+ΓΓ+ΓΓ+Γ
+
++ΓΓ+ΓΓ+
+=
321
)(
321
)5(23)3(2)(
0
0
)5(22)3()(
0
0
Λ
Λ
θβθβθβ
θβθβθβ
djsL
djsL
djLs
s
djsL
djsL
djs
s
eeeVZZ
Z
eeeVZZ
ZdV
Each of the numbered terms can now be given a physical interpretation as follows:(1) TERM: ( )[ ] )(
00θβ −+ dj
ss eVZZZ represents a forward propagating wave. At the input node of thetransmission line θβ =LLd ⇒= and the voltage is thus seen to equal ( )[ ] ss VZZZ 00 + which can beinterpreted as the source voltage Vs being divided by two series impedances of sZ and oZ (voltage dividerrule). If one thinks of this wave as the one that is initially launched then the input impedance at thetransmission line equals oZ since no wave has yet reached the load and had a chance to be reflected back tothe input. Consequently, the input impedance, inin IV , is determined initially by only the characteristicimpedance of the line. This voltage is referred to as the initially launched voltage or incident voltage anddesignate it by Vinc . Therefore, ( )[ ]00 ZZZVV ssinc +=
and the forward propagating voltage at the load
end is determined by substituting d = 0 and equals θjinceV − .
(1′) TERM: ( )[ ] )()(00
θβθβ +−+− Γ=Γ+ djLinc
djLss eVeVZZZ represents a reverse propagating wave. At
the load, d = 0 the voltage is the same as the initial wave multiplied by the load reflection coefficient LΓ .
This wave results in a voltage at the input node, l=d or θβ =l , equal to θ2jLinc eV −Γ
(2) TERM: ( )[ ] )3()3(00
θβθβ −− ΓΓ=ΓΓ+ djsLinc
djsLss eVeVZZZ represents a forward traveling wave
generated when wave 1' interacts with the source end of the line resulting in a reflected wave determined bythe source reflection coefficient.
(2′) TERM: ( )[ ] )3(2)3(200
θβθβ +−+− ΓΓ=ΓΓ+ djsLinc
djsLss eVeVZZZ represents a reverse traveling wave
resulting when wave 2 reaches the load end of the line.
The process of an incident wave resulting in repeated reflections can be represented by thefollowing diagram:
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
Thus, the voltage expression above can be seen physically to be the steady state situation resulting frommultiple reflected voltage waves on the line between the source and the load. The reflections occur becauseKirchoff's Laws are required to be satisfied at both ends of the line.
Example 2.2.1.. What is the voltage if 0ZZ L = , i.e., for a matched load?
In this case 0L =Γ and substitution yields,
)()( θβ −
+= dj
sos
o eVZZ
ZdV
The initially launched wave Vinc has a magnitude of ( )[ ]00 ZZZVV ssinc += . The phase term -θ appearssince the phase at the load ( )0=d since the load voltage will be out of phase by the electrical length of theline.
Example 2.2.2.. What is the voltage on the line if os ZZ = ?
In this case 0=Γs and
( )( ))()(
21
21
)(
θβθβ
ββθ
+−−
−−
Γ+=
Γ+=
djL
djs
djL
djjs
eeV
eeeVdV
There is an initially launched voltage of magnitude 2sinc VV = since the source voltage initially splitsevenly between the source impedance and the line characteristic impedance (which is the input impedance
0
θ
2θ
3θ
4θ
5θ
0
θ
2θ
3θ
4θ
5θ
Electrical Lenght = θ
d
Phaseor
Time
Vinc =Z o
Z oZ s +Vs( ) Vinc e j(βd- θ)
Vinc ΓL e -j(βd+θ)
Vinc ΓL e j(βd-3θ)ΓS
Vinc ΓL e -j(βd+3θ)ΓS2
Vinc ΓL e j(βd-5θ)ΓS2 2
Source LoadΓs ΓL
Figure 2.5 Illustration of multiple reflections generated by a source andload connected by a transmission line where incV = theinitially launched voltage, and θ = the electrical lengthof the line.
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
for the initial wave). A second wave results since the load is not matched. The reflected wave at the load( )0=d equals the incident wave multiplied by LΓ . The reflected voltage has a term )( θβ +− dje whichrepresents a wave propagating to the left. This is the same as the forward propagating exponential with "d"replaced by "-d." The phase delay, θ, at d=0 is the same for both exponential terms so the phase delay forthe reflected signal continues to account for the total travel delay from the source.
It is often just as easy to analyze a transmission line circuit applying the specific boundaryconditions and solving directly for the desired current or voltage. This approach often results in a betterunderstanding of the physical behavior of the circuit and is illustrated in the next set of examples.Obviously, the results must agree with those obtained by direct substitution into the above equations.
Example 2.2.3.. What is the voltage on the line if 0=sZ ?
)()( djL
dj eeAdV ββ −+ Γ+=
0/)()( ZeeAdI djL
dj ββ −+ Γ−=
sj
Lj VeeALV =Γ+= −+ )()( θθ
)( θθ jL
js eeVA −+ Γ+=
)()()( θθββ jL
jdjL
djs eeeeVdV −+−+ Γ+Γ+=
[ ]0)()()( ZeeeedI djL
djdjL
dj ββββ −+−+ Γ+Γ−=
Example 2.2.4.. Find the voltage transfer function ,(load voltage/input voltage) and input impedance for aquarter-wave transmission line.
Referring to Figure 2.5 one sees that the total voltage at a distance "d" is ( ) ( )djL
dj eeAdV ββ −Γ+= , and
therefore at the input the total voltage is ( ) ( )2/2/4/ ππλ jL
j eeAV −Γ+= ( )Ljj Γ−= which simplifies to( ) ( )LjAV Γ−= 14/λ . At the load 0=d and the total voltage equals )1()0( LAV Γ+= . The voltage
transfer ratio is found by diving V(0) by V(λ/4) which simplifies to the results in Equation (2.24). This is
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
L
LjV
VΓ−Γ+−=
11
)4/()0(
λ
o
L
ZZ
jV
V −=)4/(
)0(λ
(2.24)
a very useful result which is easy to remember since it is a simple ratio of the load impedance and thecharacteristic impedance. The -j term accounts for the phase shift naturally associated with a λ/4 line. Theload impedance can be remembered as being in the numerator since for a short circuit load the voltage ratiomust be zero.
The input impedance can be found by additionally considering the total current at an arbitrarydistance "d" from the load, 0)()( ZeeAdI dj
Ldj ββ −Γ−=
and specifically at the input where d=λ/4 the
total current is 02/2/ )()4/( ZeeAI j
Lj ππλ −Γ−=
0)1( ZA LΓ+= and therefore ,the impedance is givenby the ratio of total voltage to total current which simplifies to give the results in Equation (2.25)
oL
Lin Z
IV
ZΓ+Γ−==
11
)4/()4/(
λλ
L
oin Z
ZZ
2
= (2.25)
Power in a transmission line circuit is computed using the familiar expression .ReRe ** VIIVPower ⋅=⋅= This represents actual power transmitted or lost at particular terminals of
a circuit. The imaginary component, i.e., *Im VI , is called reactive power and represents power storedand exchanged (within a cycle) between magnetic and electric fields of the circuit. The time average ofreactive power is zero. The expression above assumes that I and V have been defined as rms currents andvoltages, which is an assumption which will always be made unless explicitly stated otherwise.
However, for transmission line circuits there are additional forms of power. Since voltage andcurrent waves can exist on a line they represent a flow of power. The power associated with these waves isfound by applying the above formula to the separate propagating components. To illustrate this considerthe figure below.
d
(Z )0
I( )d
V( )d
Z S
VS ZL
Figure 2.5 Power on the line is given by P(d)= ( )dP + - ( )dP − .
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The total voltage and current as seen earlier equals )()( djL
dj eeAdV ββ −Γ+= and
0)()( ZeeAdI djL
dj ββ −Γ−= where ( ) ( )[ ] ssj
sLj VZZZeeA 00
121 +ΓΓ−= −−− θθ . The forward voltage+V , and current +I are given by djAeV β=+ and 0ZAeI djβ=+ . The power associated with them is
the forward propagating power +P and is given by Equation (2.26a). Similarly, the reverse propagatingpower P− is given by Equation (2.26b). The net power or power delivered down the line is
−+ −= PPPdel which is shown in Equation (2.26c)
( )o
dj
o
dj
Z
Ae
ZA
AeIVP2**
ReRe =
=
= −+++ ββ
(2.26a)
oL Z
AP
22Γ=− (2.26b)
( )o
Ldel Z
AP
221 Γ−= (2.26c)
Example 2.2.5.. Compute the power using )(I(d)* dV• and compare with the previous results.
The power associated with abstract terminals at a distance "d" along the line, designated P(d), isgiven by )()(Re)( * dVdIdP = . Substitution of the expressions for I(d) and V(d) yields the results ofEquation (2.27), where .IRL jΓ+Γ=Γ .
( )( )djL
djdjL
dj
oeeee
ZA
dV ββββ −− Γ+Γ−=⋅ *2
* )(I(d)
[ ]djL
djLL
oee
ZA ββ 2*22
2
)1( Γ+Γ−Γ−= −
( )[ ]ddjZA
dV RILo
ββ 2sin2cos2)1()(I(d) 22
* Γ+Γ−Γ−=⋅ (2.27)
Taking the real part of Equation (2.27) results in ( ) 022 1)( ZAdP LΓ−= which is the net power
or power delivered by the line, delP , as seen above.
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
Example 2.2.6... Calculate the forward and reverse propagating power and the power delivered to theload for the circuit shown in Figure 2.6
For this circuit substitution into the formulas yields 103
jA −= , therefore, wattsP5000
950103 2
=
=+ .
Since 3/1=ΓL then 5000
15000
9912 =
=Γ= +− PP L . The power delivered to the load is
wattsPPPdel 6251
50001
50009 =−=−= −+
The Thevenin Equivalent Theorem applies for transmission line circuits and the equivalentvoltage and impedance is found using the familiar techniques. Given two terminals of a circuit, which caninclude any points along a transmission line, the equivalent circuit has the same voltage, currentcharacteristics that would be observed by the original circuit at the terminals. The Thevenin voltage equalsthe open circuit voltage at the terminals and the Thevenin impedance is that seen by looking into theterminals (with voltage source shorted, and current sources opened). This is illustrated by the following.
Example 2.2.7... Find the Thevenin Equivalent circuit using the output terminals of the transmission linecircuit above.
Having determined the equivalent circuit compute the power delivered to the load. The TheveninEquivalent circuit is found by opening the terminals and observing the input impedance and open circuitvoltage. This situation is illustrated in Figure 2.7.
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The input impedance can be found using the quarter wave relationship derived previously to get( ) Ω=== 25100/50100/ 22
0Thev ZZ . The open circuit voltage is found using the voltage and current
relationship for a transmission line, i.e., )()( djdj eeAdV ββ −+= and 50)()( djdj eeAdI ββ −−= , since1=ΓL for an open circuit load. At the input (d = λ/4) of the line ( ) 04/ =λV and ( ) 25/4/ jAI =λ . The
current into the transmission line under these conditions would be 1/100. This determines the constant A,)4/1(jA −= and the general voltage and current expression becomes 4)()( djdj eejdV ββ −+−=
,
, and
200)()( djdj eejdI ββ −−−= . The open circuit voltageVoc =V(0) = -j21
. The Thevenin Equivalent circuit
is therefore, Figure 2.8.
The power delivered from the transmission line to the load can now be found. The voltage acrossthe load is 5/2)10025(100)2/1( jj −=+− -. The power delivered to the load is, therefore
( )watts
RV
PL
LLOAD 625
1100
5/2 22
===
which agrees with the results from above.
2.3 GRAPHICAL ANALYSIS OF TRANSMISSION LINES
The behavior of distributed circuits with respect to parameter changes is usually very important.The two parameters whose variation is of most interest is usually physical dimensions (E.G. length), andfrequency (bandwidth). Interest in the first come about because of tolerance considerations in fabricating
ZinVoc
=100Ω
=1 ( )Zo=50Ω
λ/4
I( )d
V( )d
Z S
VS
=100 Ω ZL
Figure 2.7 Determination of Thevenin Equivalent circuit for quarterwave linewith source.
VThev = -j 12
Z Thev= 25 Ω
= 100 Ω ZL
Figure 2.8 The Thevenin Equivalent circuit for thesource and transmission line.
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
the circuits and interest in the second occurs because most circuits require more than a single frequency tooperate or communicate. In the second case the range of frequencies is the bandwidth of the system. It isconvenient to examine the effects of physical dimensions and frequency together for a distributed circuitbecause the critical parameter which determines performance is electrical length which compares physicallength with wavelength. Of course, the wavelength of a propagating signal on a distributed circuit is afunction of frequency.
Significant insights are possible by considering a transmission line with a load, LZ . The loadreflection coefficient is ΓL and from Equation (2.5a) the total voltage at a distance "d" from the load equals
( )djL
dj eAedV ββ 21)( −+ Γ+= . The magnitude of this voltage is therefore, djLeAdV β21)( −Γ+= . This
magnitude is illustrated in Figure 2.9a as the vector addition of two complex numbers, 1+j0, and djLe β2−Γ .
If the distance d is increased then the total voltage V(d) varies as shown in figure 2.9b. The vectorrepresenting the complex number dj
Le β2−Γ rotates in a clockwise fashion since negative angles aremeasured in a clockwise rotation and positive angles in a counter clockwise rotation. The total voltageincreases and decreases as the tip of the vector traces out a circle or radius LΓ . The maximum total
voltage occurs when the vector djLe β2−Γ points to the right (0 degrees) and the minimum occurs when the
vector points to the left (±180 degrees or ±π radians). A rectangular plot of the total voltage is illustrated inFigure 2.10 as a function of the distance d.
1
e- βd2jΓL
1 e- βd2jΓL+
(a.)(b.)
Figure 2.8 (a.) The total voltage on a transmission line as a vector sum.(b.) Illustration showing how the total voltage changes as thedistance "d" is increased
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
The voltage variation illustrated in figure 2.10 is called a standing wave pattern. It is an amplituderesulting from the combination of counter propagating voltages which remains stationary with respect toposition on the line, and hence the name. The minimum amplitude is designated minV and occurs in a nullin the pattern. Similarly the maximum voltage amplitude is designated maxV . The ratio of the maximum tominimum voltage on the line is called the Voltage Standing Wave Ration., minmax VVVSWR = which is areal number always greater than or equal to one. To emphasize that the VSWR is a ratio it is oftennumerical reported as "n:1", (spoken " n to one"), for example an antenna as a load at the end oftransmission line may produce a VSWR on the line of 1.4:1. Often in this situation the load is referred tohas having a VSWR which really means that it produces the VSWR when connected to the transmissionline.
The maximum voltage is given by ( )LAV Γ+= 1max and while the minimum voltage is
( )LAV Γ−= 1max . The ratio of these gives the expression for the VSWR shown in Equation (2.28a). Also,one can solve for the magnitude of the reflection coefficient which is observed to be only a function of theVSWR as seen in Equation (2.28b)
L
LVSWRΓ−Γ+
=11
(2.28a)
11
+−=Γ
VSWRVSWR
L (2.28b)
In addition to the information provided by the VSWR, the locations of the nulls, which are usuallymuch sharper than the peaks, provides useful information. The physical distance, d∆ , between successivenulls occurs when the vector in figure 2.8 rotates through 180 degrees or 2π radians. This occurs when
πβ 22 =∆d and since λπβ 2= then 2λ=∆d . Therefore, the null-to-null distance equals thehalf wavelength. Based only on measurements of the total voltage magnitude the propagation wavelengthand the magnitude of the load reflection coefficient can be determined. If the frequency is known then thepropagation phase velocity, v is determined by v f= λ. Since the load reflection coefficient is a complexnumber and only its magnitude has thus far been determined it remains to see how to calculate the angle for
VminVmax
d
V(d)λ/2
d null
ZLV(d)
Figure 2.10 Plot illustrating variations in the total voltagemagnitude )(dV
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
the load reflection coefficient. Assuming that the load reflection coefficient angle is θ then LjLL e θ+Γ=Γ
and the distance to the first null is nulld then πβ jL
djL ee null −− Γ=Γ 2 and πβθ j
Ldjj
L eee nullL −− Γ=Γ 2 which
implies that πβθ −= nullL d2 and with λπβ 2= then ( )πθ 14 −= nullL d .
Example 2.3.8.. When a 50 Ω is connected to an unknown load the following plot of voltage magnitudesare observed along the line for an operating frequency of 10 GHz. Determine the load impedance and thepropagation velocity for the line.
The problem is solved by first computing the wave length, lambda, and then computing thereflection coefficient observed at the first null, G. Note that the magnitude of G is found from VSWRformula and the angle is 180 degrees (or negative real number). The veloctiy is found since both thefrequency (given) and the wavelength (calculated) are known.
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
A straight line in the complex plane is defined by dczzc 2=+ ∗∗ where c is a complex number of havingmagnitude 1 and d is a real number. The unit vector defined by the complex number c is perpendicular tothe line and the real number d is equal to the line's distance from the origin.
Proof:
Since jyxz += and jbac += then ( ) dbyaxczzc 22 =+=+ ∗∗ and dbyax =+ is a straight
line. Since 1=+= jbac then ax by+ can be viewed as the dot product (inner product) between a unit
vector $ $ $c i i= +a bx y and a position vector yx yx iir ˆˆ += and therefore cr ˆ⋅=d and therefore d is the
projected distance from the point (x,y) in the direction of c .
Complex Geometry Theorem II:
A circle in the complex plane is defined by bczzcz =−− ∗∗2 where a is a complex number and b is a
real number such that 02 >+ cb . The center of the circle is given by the complex number a and the
radius of the circle equals 2cb + .
Proof:
Adding 2c to both sides of bczzcz =−− ∗∗2 gives 222 cbcczzcz +=+−− ∗∗ and since
( )( ) 222 czczczcczzcz −=−−=+−− ∗∗∗ then 22 cbcz +=− and 2cbcz +=− which is the
equation for a circle with center located on the complex plane at the point "c" and having radius b c+ 2 .
Note that the equation requires that 02 >+ cb
Looking at the relationship between Z and Γ one can consider how the reflection coefficientchanges as the resistance alone varies. In this case ojXRZ += where R can be viewed as a variableresistance while Xo is taken as a constant. The curve in the Z-plane is a straight line as shown in figure2.12a. Using the complex geometry theorem I the equation for the constant reactance line would be
dczzc 2=+ ∗∗ where c=0+j1, or 02XjZjZ =+− ∗ . Substitution of Z from the relationship( ) ( )Γ−Γ+= 110ZZ results in
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
and clearing the denominator results in the Equation (2.29a) where the normalized reactance is given by000 ZXx = . This simplifies to give Equations (2.29c) which can be recognized from the complex
geometry theorem II to represent a circle in the Γ-plane with center given by 011 xjc += and b=-1
implies that the radius= oo xxbc 1111 22 =−+=+ . This is illustrated in figure 2.12b. Figures 2.12c
and 2.12d illustrate the set of circle in the Γ-plane resulting from a set of constant reactance lines in the Z-plane.
In a similar way one can examine what happens to the constant resistance line in Z-plane when itis mapped into the Γ-plane. A constant resistance is presented by Z R jXo= + where X varies and Ro
Xoj Z = R+jXo
Z-Plane
real
imaj
real
imaj
1
1+ jc =
radius =
imaj
1
Z-Plane
real
imaj
(a.) (b.)
(c.)(d.)
Figure 2.12 (a.) The complex impedance plane showing the trace of a constantreactance line (b.) the constant reactance curve in the Γ-plane (c.)multiple constant reactance line in Z-plane (d.) multiple constantreactance curves in Γ-plane
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
remains constant. This is illustrated in figure 2.13a for several different choice for the constant Roincluding several that are negative. The development will apply to both positive resistance loads as well asto negative resistance loads. Again the development begins using the Complex Geometry "Theorem Iwhich describes a constant resistance line in the Z-plane, i.e., 02RZZ =+ ∗ . Substitution of
( ) ( )Γ−Γ+= 110ZZ and r R Z0 0 0= together with simplification as before yields
( )( )∗Γ−Γ−=Γ− 11222 02 r and 2
000021 Γ+Γ−Γ−=Γ− ∗ rrrr . Combining terms and division results
in Equation (2.30) which is recognized from the complex Geometry Theorem II to be a circle
0
0
0
0
0
02
11
11 rr
rr
rr
+−=Γ
+−Γ
+−Γ ∗ (2.30)
where ( ) ( )00 11 rrb +−= and ( )00 1 rrc += and for both positive and negative values of 0r one sees that
02 >+ cb . The center of the circle is therefore given by ( )00 1 rrc += and radius = 011 r+ . Note that
for 10 −>r that Center + Radius =1 i.e., the right hand side of the circles are tangent to point Γ=1+j0. For10 −<r then Center - Radius=1 and the left hand side of the circles are tangent to the point Γ=1+j0. If
10 −=r then from ( ) 20000 11 Γ++Γ−Γ−=− ∗ rrrr it follows that 2=Γ+Γ ∗ the curve is a straight line
perpendicular to the real axis and crossing the real axis at the point Γ=1. This is illustrated in figure 2.13b.
Z = R +jX
Z-Plane
real
imaj
real
imaj
1
r0 = -1
r0 = 0 r0 = 1
r0 < 0-1 <
r0 = -2
r0-2 <
r0 = -3
r0 = -1
r0 < -1-2 <
r0 < 0-1 <
r0 = 0
r0 = 1r0 = -2
r0 = -3
(a.) (b.)
Figure 2.13 (a.) Example of seven constant resistance line in the S-plane. (b.) theseven constant resistance line in the Γ-plane.
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
From the above analysis one can plot both the constant reactance and constant resistance circles onthe Γ-plane as illustrated in figure 2.14a. The dark circle equates to a resistance of zero. Inside this circlethe resistance is positive and outside of it the resistance is negative. Inside the reflection coefficient has amagnitude less than one and outside its magnitude is greater than one. The dark circleis the unit circle onthe complex Γ-plane since 1=Γ . This representation of the Γ-plane with the circles is referred to as aSmith Chart named for its founder. The region in the dark circle is referred to as the Unit Smith Chart(USC) and represents passive loads, i.e., those having an impedance with a positive real part. The regionoutside of the USC equates to active load, i.e., those having an impedance with a negative real part. TheUSC region is shown expanded in figure 2.14b
b = 1= 2 = .5bb
= .5g
= 1g
= 2g
= -2b= -1b
= -.5b
Admittance
ImpedanceCoordinates
Coordinates
(a.) (b.)
Figure 2.15 (a.) The USC showing circles of constant conductance and constantsuseptance. (b.) A Y-Z Smith Chart shows both admittance andimpedance circles
real
imaj
x = 1
x = -1
x = 2
x = -2
x = .5
x = -.5
r = 0
r = 1 r = 2r = .5
real
imaj
(a.) (b.)
Figure 2.14 (a.) The Γ-plane with the circles of constant reactance and constant resistancedisplayed. (b.) The Passive Smith Chart or Unit Smith Chart (USC)
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
If one looks at the reflection coefficient in terms of an admittance ZY 1= then the reflectioncoefficient becomes ( ) ( )00 YYYY +−−=Γ which is of exactly the same form as the resistance formulaexcept with a negative sign in the front. Therefore it follows immediately that a mapping from the complexadmittance plane would result in lines of constant conductance and constant suseptance become circles onthe Γ-plane. However the negative sign would mean that everything would be rotated by 180 degrees. Thisis illustrated in figure 2.15a for the USC where g represent conductance and b represents suseptance.Figure 2.15b illustrates the USC showing some of the admittance coordinates as well as the impedancecoordinates. This type of representation is called a Y-Z Smith Chart. Note that in all charts the bottomrepresents a capacitive load and the top an inductive load
One now considers a series resonant circuit consisting of an inductor, capacitor, and resistor as aload and considers how the reflection coefficient would change as the frequency is increased from a lowvalue to a high value. The reactance values will change from a net negative value to a net positive value asthe frequency increases. The resistor, assuming that it is ideal, will have a value independent of frequency,i.e., it will be a constant. Hence the reflection coefficient, Γ, will follow a trace similar to that illustrated inFigure 2.16a. It will lie upon a circle of constant resistance an move from capacitive to inductive reactance,i.e., move from the lower half of the USC to the upper half. The real and imaginary axis are shown as areminder that the Smith Chart is an overlay of impedance and admittance coordinates on the complex Γ-plane. Similarly, a shunt or parallel resonant consisting of an inductor, capacitor, and resistance will followa constant suseptance trace moving from inductive which will dominate a lower frequencies to capacitivewhich will dominate at higher frequencies as illustrated in Figure 2.16b.
real
imajΓ
real
imajΓ
(a.) (b.)
real
imajΓ
Lower
ResistanceShunt
(c.)
Figure 2.16 (a.) Plot of Γ with increasing frequency of a series resonant circuit, (b.) shunt or parallelresonant circuit with increasing frequency. (c.) Plot of the reflection coefficient of a shuntresonance circuit with a lower shunt resistance.
For the shunt resonator if the resistor is lower (conductance higher) then the reflection coefficientcould follow a trace similar to that shown in Figure 2.16c. Therefore, it is clear that a shunt resonatoralways follows a path from top to bottom (inductive to capacitive) and is at resonance at the frequencywhere Γ crosses the real axis. The opposite is true for a series resonance.
In figure 2.16c the Q of the resonance is lower than the previous shunt circuit. Hence the radius ofcurvature of the trace gives an indication of the Q associated with the circuit's resonance. Often a circuitmanifests more complicated behavior revealing multiple resonances. Such an example is illustrated inFigure 2.17a. In this example as the frequency is increased from a low to a higher value the circuitsreflection coefficient (or impedance) first looks capacitive and then passes through a series resonance. Asthe frequency continues to increase the circuit looks inductive and then passes through a shunt resonance.
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
As the frequency is increased even higher the circuit passes through a series resonance again andremains inductive thereafter. The example of figure 2.17a illustrates the property that loops in the reflectioncoefficient trace proceed in a clockwise fashion as frequency increases. This usually makes it possible toexamine data on a Smith Chart and deduce which point on the trace is the lower versus the higherfrequency. This behavior of the reflection coefficient trace is due to of causality property of a physicallyrealizable circuit. This follows from the fact that the reflection coefficient as a function of frequency isrelates an reflected voltage to an incident voltage by )()()( ωωω +− Γ= VV . If the incident voltage is
viewed as a time signal whose Fourier Transform is )(ω+V then ∫+ ∞
∞−
++ = ωωπ
ω deVtv tj)(21
)( . Likewise the
reflected time signal would be related to )(ω−V by its Fourier transform, i.e.,
∫+ ∞
∞−
−− = ωωπ
ω deVtv tj)(21
)( . Also )()()( tvtgtv +− ⊗= where ⊗ represents the convolution operation, i.e.,
∫+ ∞
∞−
+− −= τττ dgtvtv )()()( . If the circuit is a physically realizable one then the principle of causality
applies. This means that the signal output at a time "t" can only depend of signal inputs occurring onlybefore time "t" and not afterwards. Therefore from the integral it must follow that g(τ) = 0 when τ<0 sinceotherwise the integral would have a contribution from )(tv + for times exceeding t. Also,
dtetg tjωω −+ ∞
∞−∫=Γ )()( and since g(t)=0 for negative times then dtetg tjωω −
+ ∞
∫=Γ0
)()( . Notice that Γ( )ω is a
sum (integral) of complex unit vectors, e j t− ω , rotating clockwise as a function of increasing frequency, ω,and which are scaled by an amplitude factor, g(t). Therefore a trace of Γ( )ω can be though of as a vectorsum as illustrated in figure 2.17b.
real
imaj
sum
Γ-Plane
real
imaj
(a.) (b.)
g(t )e1-jω t11
g(t )e2-jω t21
g(t )e3-jω t31
g(t )e1-jω t12
g(t )e2-jω t22
g(t )e3-jω t32
Figure 2.17 (a.) More complex circuits display multiple resonant behavior (2series+1 shunt). (b.) Tendency towards clockwise rotationdue to causality of physical circuits
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
Figure 2.18 illustrates a Γ plot of a transmission line with a load as a function of frequency. Theradius of the circle is LΓ . Again notice that the trace proceeds in a clockwise direction with increasingfrequency.
real
imajΓ Z0( ) ZL
ΓL
Figure 2.18 Reflection coefficient trace as a function of frequency for atransmission line connected to load impedance
Example 2.3.1.. Find the reflection coefficient for a load consisting of a series combination 75 Ωresistance and 1 pF capacitor operating a 1 GHz where Zo= 50 Ω . What is reflection coefficient the loadis connected to a λ/2 line, λ/4 line, or λ/8 line.
The calculations are easily computed using the above dirived relationships
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
Example 2.3.2.. A new load is created by making parallel connection of two 50 Ω lines each having theirown load as illustrated below. What is the impedance. What is the reflection coefficient produced by thenew load on a 75 Ω line with length λ/4.
ZL2=50+j*25;GL2=(ZL2-Zo)/(ZL2+Zo)GL2 = 0.0588 + 0.2353iG2=GL2*exp(-j*2*pi)G2 = 0.0588 + 0.2353i» ZL2p=Zo*(1+G2)/(1-G2)ZL2p = 50.0000 +25.0000i [Note same as ZL2 since line is lambda/2]
ZL=(ZL1p*ZL2p)/(ZL1p+ZL2p) [Parallel combination]
ZL = 22.4138 -18.9655i
Z1=75; [Characteristic impedance of line]GL=(ZL-Z1)/(ZL+Z1)
GL = -0.4836 - 0.2888i
Example 2.3.3.. For the load illustrated below compute the reflection coefficient as the frequencyincreases from .5 to 8 GHz..
YL=Y1+Y2+Y3;ZL=1./YL;GL=(ZL-Zo)./(ZL+Zo);G=GL.*exp(-j*2*beta*length);plot(G)hold onusc1axis offfunction usc1%over lays Unit smith chart on ploted complex dataplot(res2usc(0))hold on
(Z )0
LC1
L1
R1
C2
R2Γ C1
L1
R1
C2
R2
=
=
=
=
=
Z 0 = 50 Ω1 nH
500 Ω
.5 pF
2 Ω
10 pFL = .05 cm
v = 2 x 1010 cm/s
Figure 2.20 VSWR data for Example 2.3.3
#1 #2
#1 = Series Resonance#2 = Shunt or Parallel Resonance
f1 = .5 GHzf2 = 8 GHz
Figure 2.21 Reflection coefficient trace for thecircuit in Example 2.3.3
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
axis('square')plot(res2usc(1),'c--') %plot(res2usc(1/3),'c--') Comment out unwanted coordinates %plot(res2usc(.5)) %plot(res2usc(.25)) %plot(res2usc(2)) %plot(res2usc(3),'c--') %plot(res2usc(4))plot(x2usc(0),'c:')plot(x2usc(1),'c:') %plot(x2usc(3),'c:') %plot(x2usc(1/3),'c:') %plot(x2usc(.5)) %plot(x2usc(.25)) %plot(x2usc(2)) %plot(x2usc(4))plot(x2usc(-1),'c:') %plot(x2usc(-3),'c:') %plot(x2usc(-1/3),'c:') %plot(x2usc(-.5)) %plot(x2usc(-.25)) %plot(x2usc(-2)) %plot(x2usc(-4))--------------------------------function y=res2usc(r)%generates constant resistance circle for Unit Smith Charttheta=pi*[0:5:360]/180;Ucir=exp(j*theta);ctr=r/(1+r);rad=1/abs(1+r);y=ctr+rad*Ucir;
-------------------------------function y=x2usc(x)%generates constance reactance circle-segments for Unit Smith Chartif x==0 y=[-1+j*eps, 1+j*eps];else ctr=1+j/x; rad=1/abs(x); maxangle=2*atan(x); if x>0 theta=-pi/2-[0:maxangle/20:maxangle]; elseif x<0 theta=pi/2-[0:maxangle/20:maxangle]; end y=ctr+rad*exp(j*theta);end
2.4 TRANSMISSION LINE STUBS (λ/4. etc.)
Transmission lines where the load is either an open circuit or a short circuit are of special importance. Inthe first case ∞== openL ZZ and the second Z ZL short= = 0 , and the reflection coefficient becomes Γopen = 1
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
and Γshort = − 1 . The total voltage and currents at a location "d" from such loads are given byV d A e eopen
j d j d( ) ( )= ++ −β β , 0/)()( ZeeAdI djdjopen
ββ −+ −= , and V d A e eshortj d j d( ) ( )= −+ −β β ,
0/)()( ZeeAdI djdjshort
ββ −+ += . Each of these expression can be simplified using Euler's identity to getdAdVopen βcos2)( = , 0/sin2)( ZdAjdIopen β= , and dAjdVshort βsin2)( = , 0/cos2)( ZdAdI short β= .
The impedance looking through a line of length d at an open or short load is given by)()()( dIdVdZ openopenopen = and )()()( dIdVdZ shortshortshort = resulting in Equations 2.31a and 2.31b.
This is illustrated in figure 2.22 which shows that the VSWR = ∞ Open and short circuited transmission lineare often referred to an open stub or shorted stub.
djZdZopen βcot)( 0−= (2.31a)
djZdZ short βtan)( 0= (2.31b)
d
V(d)
I(d)
d
V(d)
I(d)
d d
X (d)open X (d)short
Z (d) = jX (d)open open Z (d) = jX (d)short short
Impedance Impedance(a.) (b.)
Figure 2.22 (a.) )(dV , )(dI , Reactance plots, and impedance type for an opencircuited transmission line. (b.) same for a short circuit transmission line.
The reflection coefficient for an open circuited stub is dje β2−=Γ and therefore data as a function oflength, "d," or as a function of increasing frequency would produce a trace on the unit circle of Smith Chartdisplay. This is illustrated in figure 2.23 which shows the impedance type associated with the point.
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
openshort
Figure 2.23 A reflection coefficient trace of an open stub on the USC followsthe unit circle rotating in a clockwise direction starting fromthe right (open) value. Trace is a function of increasing lengthor increasing frequency.
Example 2.4.1.. A circuit consisting of two sources, one operating at 1 GHz and other at 2 ±.05 GHz areconnected via a 50 Ω resistor to a transmission line which in turn is connected to a 50 Ω load. It is desiredto suppress the 2 GHz signal using an open circuited stub. If a 2 GHz λ/4 open stub is connected in shuntwith the load what suppression can be achieved as a worst case.? How does the characteristic impedance ofthe stub affect the power delivered to the load at the desired frequency?
This problem is easily examined using computer aided analysis. The solution below assumes a propagationvelocity of 20 cm/ns for the direct line and stub (all impedances). A frequency of 2 GHz equates to awavelength λ/4=5 cm which becomes the length for the stub, D1.
Pmin=-80*ones(size(Parray));plot(f*1e-9,max(Parray,Pmin))axis([1 2.5 -80 0 ])title('Power delivered to load vs Frequency')xlabel('Frequency (GHz)')ylabel('Power (dBm)')Compare=[Parray(:,find(f==1e9)),Parray(:,find(f==1.95e9))];Compare(:,1)-Compare(:,2)
20 25.9702
80 17.6748
Example 2.4.2.. A λ/4 stub is used so that the voltage from a DC source (called a bias voltage) can beapplied to the output of a transistor. Find the percentage of power that goes to the laod vs. the bias supply.for the ciruit below.
It is desirable for no DC power to reach the load so a series capacitor is used. It is also desireable for verylittle microwave power to propagate down the stub into the bias circuits so the source should appear as alow impedance.
1 1.5 2 2.5-80
-70
-60
-50
-40
-30
-20
-10
0Power delivered to load vs Frequency
Frequency (GHz)
Power (dBm)
Z stub=205080
110
Figure 2.25 Plot showing power delivered to load as afunction of frequency for a fixed stub
Power to load at desired freq compared to power to load atunwanted freq
Zcomb=Zstub*Z2/(Zstub+Z2);Gcomb=(Zcomb-Zo)/(Zcomb+Zo);Gin=Gcomb*exp(-j*2*pi*theta1);Zin=Zo*(1+Gin)/(1-Gin);Iin=V./(Rtran+Zin);Vin=V*Zin./(Rtran+Zin);P=conj(Iin).*Vin;Pin=real(P);Pcomb=Pin;V2magsqr=Pcomb*Zcomb;Pstub=real(V2magsqr/Zstub);P2=real(V2magsqr/Z2);Loadfraction=10*log10(P2/Pin) % dBStubfraction=10*log10(Pstub/Pin) % dB
To Load -0.3135 dB
(Z )0
A
B(Z )0
(Z )1
C
Transistor
Bias5 v
1 Ω
50 pF
R = 50L Ω
λ/4
λ/8 λ/163 7
Figure 2.26 Plot showing power delivered toload as a function of frequency for a fixed stub
Fractional Power split between Load and Bias
Where Power Delivered Power Ratio (dB)To Load -0.3135 dB
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
2.5 LOSSY TRANSMISSION LINES
In some cases it is desireable to consider the losses associated with propagation on transmissionlines. This may occur when circuits involve longer lines, or for lines required to be fabricated with lossymaterials, or for circuits operating at very high frequencies such as millimeter wave circuits. For a low losstransmission line, it is generally assumed that LR ω<< and CG ω<< . With these assumptions one candetermine the characteristic impedance cZ the phase velocity of the signal using the expression for thepropagation constant γ derived in section 2.1. The low loss assumptionsimply that
CLZ c ≅ (2.32)
which is the same as the characteristic impedance of a lossless transmission line. For the propagationconstant, one should proceed as follows. From (2.9),
( )LGRCjLCRG ++−= ωωγ 2 (2.33)
With the low loss assumption, one can dismissed the RG term in the equation above since it is too smallcompared to other terms. Thus,
( )LGRCjLC ++−≅ ωωγ 2 (2.34)
Applying the binomial expansion, i.e. ( ) ( ) ⋅⋅⋅+−++=− −− 221
!21
xann
xnaaxa nnnn , to (2.34).
( ) ( ) ( ) ( ) ( ) ⋅⋅⋅++−++−+−≅ −− 2223
221
221
2
81
21
LGRCLCLGRCjLCLC ωωωωωγ
or
⋅⋅⋅+
++
++≅
2
821
CG
LRLC
jCG
LRLCLCj
ωωωωγ (2.35)
All terms after the second in the above equation is negligible based on the small loss assumptions. And,finally,
LCjCG
LR
LCj ωβαγ +
+=+≅
21
(2.36)
The phase constant β is also the same as the wave number defined in a lossless transmission line, but nowthe attenuation constant is not zero. Since the wave is attenuated by the factor of xe α− as it traveled in the
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
The complete solution of the voltage wave that travels in the x+ direction is tjxeV ωγ +−+ . The phasevelocity is derived from the time displacement of the constant phase point, i.e. assuming the exponentialterm to be a constant, utjx =+− ωγ , and taking the time derivative of it. This yields
LCLCC
GL
RLC
jLCj
CG
LR
LC
jjtdxd
v p1
21
1
21
≅+
+
=+
+
≅==
ωωω
ωγω (2.38)
The phase velocity of the signal in a low loss line is the same as that in a lossless line. It is customary toexpressed the characteristic impedance in terms of the phase velocity and one of the circuit parameter asbelow.
CvLv
CLZ
ppc
1=== (2.39)
At high frequencies, the electromagnetic field can penetrate only a small distance into a conductor.In fact, the amplitude of the fields decay exponentially from its value at the surface of the conductoraccording to sue δ− . Here, u is the normal distance into the conductor and sδ is the skin depth givenbelow.
mms σµω
δ 2= (2.40)
where ω is the angular frequency of the signal, mµ is the permeability of the conductor, and mσ is theconductivity of the conductor. At one skin depth, i.e. su δ= , the field strength is 37% of the surface value.At three skin depth, i.e. su δ3= , the field strength is down to only 5% of its surface value. For copper, acommonly used conductor in microwave transmission media, the skin depth at 100 MHz is 6.6 micron or0.26 mils. At 10 GHz, the skin depth is 10 times smaller than that. Therefore, when considering the finiteconductance of a lossy transmission line, the series resistance R in Figure 2.1 is usually expressed in termsof the surface resistance Rm , which is defined as
smmR
δσ1= (2.41)
The exact formula for the series resistance of a particular transmission media also depends on the geometryof the media itself.
In Figure 2.1, the admittance Y of the shunt components can be expressed as
−=+=
CGjCjGCjY
ωωω 1
The term CG ω , which causes the admittance Y to deviate from a pure imaginary value with non-zerodielectric conductivity, is called the loss tangent and conventionally denoted as " tanδ". For a particulartransmission media, the geometric factor for the conductance and capacitance of the dielectric is the sameand can be eliminated from the loss tangent formula. In other words, dkG σ= and dkC ε= , where k is the
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
geometric factor, dσ and dε is the conductivity and the permittivity of the dielectric respectively.Therefore,
d
d
εωσδ=tan (2.42)
and the dielectric conductance G in terms of the loss tangent is
δω tanCG = (2.43)
To account for attenuation due transmission line losses is reatively easy to accomadate since γ = α + jβ andthe total voltage on the line is
( )dL
d eeAdV γγ −Γ+=)(
The generalized reflection coefficient at a point "d" from the load is
dLe γ2−Γ=Γ
The attenuation constant has units of inverse distance since when it is multiplied by "d" the exponentbecomes a unitless number. This unitless number is given the honorary label, Neper, after the Napier theinventor of natural logarithms. Thus, α would be described in terms of Nepers/cm, Nepers/in indicatingthat when multiplied by the length in appropriate units that the resulting number is the correct exponentialvalue. Howeve, loss is usually measured in terms of dB or dB per unit length and it is therefore importantto understand how to convert to Nepers per unit length for calculation purposes. The voltage a point "d" ona matched line is compared with the voltage at a point "d+∆d" to get
djddd
ddeee
AeAe
dVddV ∆∆∆
∆+===∆+ βαγ
γ
γ )(
)()(
dedV
ddVdP
ddP ∆=∆+=∆+ α22
)()(
)()(
eddP
ddPPdB 1010 log20
)()(
log10 ∆=
∆+=∆ α
686.8log20 10 ==∆
∆e
dPdB α
Therefore 1 Neper/unit length is seen to equal 8.68 dB/unit length. It is instructive to examine a line thathas 1 dB of loss per wavelength.
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
Example 2.5.1.. Plot the total voltage magnitude on a rectangular plot and the reflection coefficient on aSmith chart for an open circuited stub with a 1dB per wavelength loss.
%script file ex2_4_1%to illustrate attenuation
% alpha is 1 dB / wavelength% beta is just 2*pi radians since% d is measured as fractional% wavelenghtd=[0:.01:3];beta=2*pi;alpha=1/8.686;gamma=alpha+j*beta;GL=1;Vd=exp(gamma*d)+exp(-gamma*d);Vdmag=abs(Vd);plot(d,Vdmag)
G=exp(-2*gamma*d);figureplot(G,'r');hold
Figure 2.27 Plot showing reflection coefficient for loss opencircuit stub for lengths up to 3 λ.
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
usc
From Figure 2.28 it is noted that the VSWR changes as the measurement moves away from the open end ofthe stub. Measurements of the VSWR at multiple cycles from the open can be used to detemine the lossassociated with the line. From the smith chart display one see that the trace makes 6 cycles equating to alength of 3λ. One dB per wavelength causes the reflection coefficient to spiral inward so that its radius hasreduced from 1 to 1/2 for an open circuited stub.
00.511.522.530
0.5
1
1.5
2
2.5
(Z )0 Loss = 1 dB/wavelength open
d (fractions of wavelength)
Figure 2.28 Rectangular plot of total voltage for loss open circuit stubwith lengths up to 3 λ.
Chap_02r.doc All Rights Reserved -- M. L. Edwards, 7 September, 2001
2.6 PROBLEMS
1. The function y=cos(ωt-βx) reprsents a propagating wave on the x-axis. If the frequency is 1 GHz andthe wave-length is 20 cm plot y vs x for t=0, .2ns, .4ns, .6ns where the x-axis goes from 0 to 100 cm.Measure the movement of a wave crest for the different time intervals and calculate the phase velocity.Compare with the formula v=fλ.
2. A lossless transmission line has a distributive inductance of 1 nH/cm, a distributive capacitance of .5pF/cm. What is the characteristic impedance of the line. What is the phase velocity. If a wave on this linehas a frequency of 1 GHz what is its wavelength?
3. For the circuit shown use Kirchoff's voltage law (KVL) to write two equations in terms of the currents I1
and I2 . Express the equations as a matrix equation BII
A2
1 =
where A is a 2x2 matrix and B is a 2x1
column matrix. Solve for the current matrix by inverting A. What is the power delivered to RL?
Vs
Rs L
CRL
VsRs
LCRL
f ===
===
1 GHz1 volt50 Ω 50 Ω
5 nH7 pFI1 I2
VA VB
4. For the circuit shown in problem 3 use Kirchoff's current law (KCL) to write two equations in terms of
the voltages V1 and V2 . Express the equations as a matrix equation DVV
C2
1 =
where C is a 2x2 matrix
and D is a 2x1 column matrix. Solve for the current matrix by inverting C. What is the power delivered toRL?
5. Solve the circuit in problem 3 by determining the Thevenin Equivalent for the circuit below an thenconnect the resistor RL and determine the power delivered to it.
Vs
Rs L
CRLVT
Z T
6. What is the power delivered to each of the the loads R1 and R2 ?
λ/8 λ/4
50 Ω
25 Ω 100Ω( ) ( )R1
R2
R1
R2
=
= 50 Ω
100 Ω1 v
7. For an arbitary length, "L," of transmission line find the voltage transfer ratio (total voltage at loaddivided by total voltage at input. Express ratio in terms of electrical length, θ=βL. Substitute special casewhere L=λ/4 and demonstrate results in chapter.
All Rights Reserved -- M. L. Edwards, 7 September, 2001 Chap_02r.doc
Z inSource Load
x d
L
(Z )0
I( )d
V( )d ZL
θ = β L
8. Plot the reflection coefficient with respect to Zo=50 Ω on the Unit Smith Chart for a frequency rangefrom .1 GHz to 40 GHz. Identify the resonance points and characterize their type.
